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3 years ago

14

Arabella's password consists of 1 letter followed by two digits. What is the probability of correctly guessing Arabella's passwo

rd?

1 answer:

Alchen [17]3 years ago

6 0

The answer is 1/2600

Hope this helps

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What's 19/30 x 12/16?

mr Goodwill [35]

19/40 as a fraction and decimal it is 0.475

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3 years ago

Read 2 more answers

Least common denominator in 1/30 + 1/24

aev [14]

Answer:

The LCD in 1/30 + 1/24 is <u>90</u>.

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2 years ago

Suppose that you draw two cards from a deck. After drawing the first card, you do not put the first card back in the deck. What

AVprozaik [17]

Answer:

The correct answer is 0.05882.

Step-by-step explanation:

A deck of cards have 52 cards, 13 cards of each suit.

We are drawing two cards without replacement.

We need to find the probability of getting a diamond as the first card.

Favorable outcomes are 13 and total number of outcomes are 52.

Thus this probability is $\frac{13}{52} = \frac{1}{4}$ .

Now for the next draw we again want to pick a diamond card.

Favorable outcomes are 12 and total number of leftover cards are 51.

Thus this probability is $\frac{12}{51}$ .

Now the probability that both cards are diamonds is $\frac{1}{4}$ × $\frac{12}{51}$ = $\frac{3}{51} = \frac{1}{17}$ = 0.0588235 ≈ 0.05882

4 0

3 years ago

A hockey team is convinced that the coin used to determine the order of play is weighted. The team captain steals this special c

fredd [130]

Answer:

Since x= 12 (0.006461) does not fall in the critical region so we accept our null hypothesis and conclude that the coin is fair.

Step-by-step explanation:

Let p be the probability of heads in a single toss of the coin. Then our null hypothesis that the coin is fair will be formulated as

H0 :p 0.5 against Ha: p ≠ 0.5

The significance level is approximately 0.05

The test statistic to be used is number of heads x.

Critical Region: First we compute the probabilities associated with X the number of heads using the binomial distribution

Heads (x) Probability (X=x) Cumulative Decumulative

0 1/16384 (1) 0.000061 0.000061

1 1/16384 (14) 0.00085 0.000911

2 1/16384 (91) 0.00555 0.006461

3 1/16384(364) 0.02222

4 1/16384(1001) 0.0611

5 1/16384(2002) 0.122188

6 1/16384(3003) 0.1833

7 1/16384(3432) 0.2095

8 1/16384(3003) 0.1833

9 1/16384(2002) 0.122188

10 1/16384(1001) 0.0611

11 1/16384(364) 0.02222

12 1/16384(91) 0.00555 0.006461

13 1/16384(14) 0.00085 0.000911

14 1/16384(1) 0.000061 0.000061

We use the cumulative and decumulative column as the critical region is composed of two portions of area ( probability) one in each tail of the distribution. If alpha = 0.05 then alpha by 2 - 0.025 ( area in each tail).

We observe that P (X≤2) = 0.006461 > 0.025

and

P ( X≥12 ) = 0.006461 > 0.025

Therefore true significance level is

∝= P (X≤0)+P ( X≥14 ) = 0.000061+0.000061= 0.000122

Hence critical region is (X≤0) and ( X≥14)

Computation x= 12

Since x= 12 (0.006461) does not fall in the critical region so we accept our null hypothesis and conclude that the coin is fair.

3 0

3 years ago

Through point (0, 10) and PERPENDICULAR to the line through (-2, -9) and (-4, -17)?

inn [45]

Compute the slope of the line through (-2, -9) and (-4, -17):

$\dfrac{-9-(-17)}{-2-(-4)}=\dfrac82=4$

Any line perpendicular to this line will have slope -1/4. The one that passes through (0, 10) satisifes

$y-10=-\dfrac14(x-0)\implies\boxed{y=10-\dfrac x4}$

3 0

3 years ago